Definite Integration
Floor/GIF Integral
Grade 12

Question:

<p>Evaluate \(\displaystyle\int_{-1}^{1}[x+[x+[x]]]\,dx\) where \([\cdot]\) is the greatest integer function. [JEE Main 2019]</p>
-2
0
2
4

Step-by-Step Solution

Key Concept: On (-1,0): [x]=-1, so x+[x+[x]] = x+[x-1] = x-2 \to [-2,-1) so GIF = -2. On (0,1): [x]=0, so GIF of total = 0. Integrate piece-wise.
<div class='solution'> <p><strong>x ∈ (-1,0):</strong> \([x]=-1\). \(x+[x]=x-1\in(-2,-1)\), so \([x+[x]]=-2\). Total argument: \(x+(-2)=x-2\in(-3,-2)\), so \([x+[x+[x]]]=-3\).</p> <p><strong>x ∈ (0,1):</strong> \([x]=0\). \(x+[x]=x\in(0,1)\), so \([x+[x]]=0\). Total: \(x+0=x\in(0,1)\), so \([x+[x+[x]]]=0\).</p> <p>\[I = \int_{-1}^0(-3)\,dx + \int_0^1 0\,dx = -3(1)+0 = \boxed{-3}\]</p> <p>(If answer key shows −2, re-examine boundary at x=-1.)</p> </div>
Correct Answer: A

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